Question

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

Short answer

The enthalpy change for the desired reaction, \(P_4O_{10}(s) + 6PCl_{5}(g) \rightarrow 10Cl_{3}PO(g)\), is \(\Delta H = 8545.8 \mathrm{kJ}\).

Step-by-step solution

Analyze the desired reaction

First, let's write the desired reaction: $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$ We will now use the given reactions and manipulate them to obtain the desired reaction.

Manipulate the given reactions

Let's label the given reactions as R1, R2, R3, and R4 as below and manipulate them: R1: \(P_{4}(s) + 6Cl_{2}(g) \longrightarrow 4PCl_{3}(g)\) \(\Delta H_1 = -1225.6\) R2: \(P_{4}(s) + 5O_{2}(g) \longrightarrow P_{4}O_{10}(s)\) \(\Delta H_2 = -2967.3\) R3: \(PCl_{3}(g)+Cl_{2}(g) \longrightarrow PCl_{5}(g)\) \(\Delta H_3 = -84.2\) R4: \(PCl_{3}(g)+\frac{1}{2} O_{2}(g) \longrightarrow Cl_{3}PO(g)\) \(\Delta H_4 = -285.7\) Manipulate reactions: Reverse R2 and multiply by 2: 2 * (−R2): \(2P_{4}O_{10}(s) \longrightarrow 2P_{4}(s) + 10O_{2}(g)\) \(\Delta H_{-2} = 2(2967.3)\) Multiply R3 by 6: 6R3: \(6PCl_{3}(g) + 6Cl_{2}(g) \longrightarrow 6PCl_{5}(g)\) \(\Delta H_{3} = 6(-84.2)\) Multiply R4 by 10: 10R4: \(10PCl_{3}(g) + 5O_{2}(g) \longrightarrow 10Cl_{3}PO(g)\) \(\Delta H_{4} = 10(-285.7)\)

Combine the reactions

Now, we will add the manipulated reactions to get the desired reaction: 2 * (-R2) + 6R3 + 10R4: \(2P_{4}O_{10}(s) + 6PCl_{5}(g) \longrightarrow 10Cl_{3}PO(g)\)

Calculate the enthalpy change

Lastly, add the enthalpy changes of the manipulated reactions to find the enthalpy change for the desired reaction: \(\Delta H = 2\Delta H_{-2} + 6\Delta H_{3} + 10\Delta H_{4}\) \(\Delta H = 2 (2 * 2967.3) + 6 (-84.2) + 10(-285.7)\) \(\Delta H = 11908 - 505.2 - 2857\) \(\Delta H = 8545.8 \mathrm{kJ}\) Therefore, the enthalpy change for the desired reaction is: \(\Delta H = 8545.8 \mathrm{kJ}\)

Key concepts

Thermochemical Equations

Thermochemical equations are not just chemical equations; they elegantly combine stoichiometry with energy changes. These equations depict both the quantity of substance involved in the reaction and the energy exchanged in the form of heat, \( Q \), during the reaction at constant pressure (also known as enthalpy change, \( \Delta H \)).

For instance, take the combustion of methane: \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)+ Q \). If the reaction releases 890.4 kJ of heat per mole of \( CH_4 \) burned, we would write \( CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \Delta H = -890.4 kJ/mol \).

Understanding thermochemical equations is critical as they tell us whether the reaction is exothermic (energy releasing, \( \Delta H < 0 \) ) or endothermic (energy absorbing, \( \Delta H > 0 \) ). This concept of enthalpy change is key in various scientific fields, from designing energy-efficient processes to understanding environmental impacts.

Hess's Law

Hess's Law is a powerful tool in chemical thermodynamics for calculating the enthalpy change of a reaction. It states that the total enthalpy change during a chemical reaction is the same no matter how many steps the reaction is carried out in. This is because enthalpy is a state function, which means its value depends only on the initial and final states, not on the path taken.

Here's a practical application: if you want to find the enthalpy change for a reaction that is difficult to measure directly, you can instead find the enthalpy changes for a series of reactions that add up to your target reaction. The sum of these changes will give you the desired enthalpy change, as seen in the step-by-step solution provided in the exercise.

Why is this useful? Sometimes, direct measurement of the enthalpy change in a reaction is impossible or impractical because of, for instance, extremely high temperatures, slow reaction rates, or the lack of appropriate experimental setup. Hess's Law gives us a detour around these roadblocks.

Enthalpy of Reaction

The enthalpy of reaction, often denoted as \( \Delta H_{reaction} \), is the heat change that occurs at constant pressure when a reaction takes place. It's calculated by summing the enthalpy of formation values for all products and then subtracting the sum of the enthalpy of formation values for all reactants.

When a reaction is exothermic, \( \Delta H_{reaction} \) is negative, meaning that the system releases heat to the surroundings. Conversely, an endothermic reaction has a positive \( \Delta H_{reaction} \) and absorbs heat from the surroundings.

The exercise provided demonstrates how you might calculate the enthalpy of a complex reaction by manipulating and combining simpler reactions with known enthalpy values. This manipulation follows Hess's Law, as we've discussed, and leads us to the calculated enthalpy change for the final reaction. Such calculations are everyday tasks in chemistry and essential in industries where energy changes need to be managed efficiently, like in pharmaceuticals and material sciences.